# limiting distribution of $Y=\nu_1X$ is chi-square with $\nu_1$ degrees of freedom

suppose $X$ has an F distribution with $\nu_1$ and $\nu_2$ degrees of freedom . it is well known that as $\nu_2$ approaches infinity. how can show limiting distribution of $Y=\nu_1X$ is chi-square with $\nu_1$ degrees of freedom

## 1 Answer

The density function of the $F$-distribution is

$$f_X(x; d_1,d_2) = \frac{1}{\mathrm{B}\!\left(\frac{d_1}{2},\frac{d_2}{2}\right)} \left(\frac{d_1}{d_2}\right)^{\frac{d_1}{2}} x^{\frac{d_1}{2} - 1} \left(1+\frac{d_1}{d_2}x\right)^{-\frac{d_1+d_2}{2}}$$

For $Y=d_1X$ the change-of variable formula immediately gives

$$f_Y(y; d_1,d_2) = \frac 1{d_1}\cdot f_X\left(\frac 1{d_1}y; d_1,d_2\right)=\\ =\frac 1{d_1}\frac{1}{\mathrm{B}\!\left(\frac{d_1}{2},\frac{d_2}{2}\right)} \left(\frac{d_1}{d_2}\right)^{\frac{d_1}{2}} \left (\frac 1{d_1}\right)^{\frac{d_1}{2} - 1} y^{\frac{d_1}{2} - 1} \left(1+\frac{d_1}{d_2}\frac 1{d_1}y\right)^{-\frac{d_1+d_2}{2}}$$

Simplifying and using the relation $\mathrm{B} (x,y)=\dfrac{\Gamma(x)\,\cdot\Gamma(y)}{\Gamma(x+y)}$

$$=\frac{\Gamma\left(\frac{d_1}{2}+\frac{d_2}{2}\right)}{\Gamma(\frac{d_1}{2})\,\cdot\Gamma(\frac{d_2}{2})} \left(\frac{1}{d_2}\right)^{\frac{d_1}{2}} y^{\frac{d_1}{2} - 1} \left(1+\frac{1}{d_2}y\right)^{-\frac{d_1+d_2}{2}}$$

and rearranging and simplifying yet more

$$f_Y(y; d_1,d_2) =\frac{\Gamma\left(\frac{d_1}{2}+\frac{d_2}{2}\right)}{\left(d_2+y\right)^{\frac{d_1}{2}}\,\cdot \Gamma(\frac{d_2}{2})} \frac 1{\Gamma(\frac{d_1}{2})} y^{\frac{d_1}{2} - 1} \left(1+\frac{1}{d_2}y\right)^{-\frac{d_2}{2}}$$

Then $$\lim_{d_2\rightarrow \infty}f_Y(y; d_1,d_2) = \lim_{d_2\rightarrow \infty}\left[\frac{\Gamma\left(\frac{d_1}{2}+\frac{d_2}{2}\right)}{\left(d_2+y\right)^{\frac{d_1}{2}}\,\cdot \Gamma(\frac{d_2}{2})}\right]\cdot\frac 1{\Gamma(\frac{d_1}{2})} y^{\frac{d_1}{2} - 1} e^{-\frac{y}{2}}$$

For this to be the density of a chi-square distribution with $d_1$ degrees of freedom, we finally need to show that

$$\lim_{d_2\rightarrow \infty}\left[\frac{\Gamma\left(\frac{d_1}{2}+\frac{d_2}{2}\right)}{\left(d_2+y\right)^{\frac{d_1}{2}}\,\cdot \Gamma(\frac{d_2}{2})}\right] = 2^{-\frac {d_1}{2}}$$

I will show it for $d_1, d_2$ even numbers. Then $m_1\equiv d_1/2,\; m_2\equiv d_2/2$ are integers. Then the limit becomes

$$\lim_{d_2\rightarrow \infty}\left[\frac{(m_1+m_2-1)!}{\left(2m_2\right)^{\frac{d_1}{2}}\,\cdot (m_2-1)!}\right] = 2^{-\frac {d_1}{2}}\cdot\lim_{d_2\rightarrow \infty}\left[\frac{m_2\cdot(m_2+1)\cdot ...\cdot (m_1+m_2-1)}{m_2^{m_1}}\right]$$

The numerator of the limit has $m_1$ terms so the leading term of the product will be $m_2^{m_1}$, the same as the denominator. So the limit tends to unity. QED.